Let K be a real number. The two real roots of the quadratic equation x2 + KX + K + 1 = 0 of X are X1 and X2 respectively. If X1 + 2 [x2] 2 = k, the value of K is obtained RTRTRTRTRT It doesn't matter. I already know the answer. Now it depends on who answers. -

Let K be a real number. The two real roots of the quadratic equation x2 + KX + K + 1 = 0 of X are X1 and X2 respectively. If X1 + 2 [x2] 2 = k, the value of K is obtained RTRTRTRTRT It doesn't matter. I already know the answer. Now it depends on who answers. -

If X2 is the root of x2 + KX + K + 1 = 0, then x2 & sup2; + kx2 + K + 1 = 0 (1) Weida theorem: X1 + x2 = - K (2) X1 + 2x2 ^ 2 = K (3) (3) - (2) has 2x2 ^ 2-x2 = 2K (4) (4) - 2 (1) has 2x2 ^ 2-x2-2x2 ^ 2-2kx2-2k-2 = 2k-x2-4k-2 = 0x2 + 2kx2 + 4K + 2 = 0 (x2 + 2) (2k + 1) = 0k = - 1 / 2 or x2 = - 2

Two second radical addition and subtraction problems in grade two of junior high school
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（1）√8/3 + √1/2 + √0.125 - √6 + √32
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（2）1/4√32a + 6a√a/18 - 3a^2√2/a
This is another question I asked,
But no one has answered,
OK, it will be added

1、=（2√6）/3+(√2)/2+(√2)/4-√6+4√2=(19√2)/4-(√6)/3
2、=√2a+6a/6*(√2a)-3a（√2a)=(1-2a)√2a

Mathematics problems in grade two of junior high school
Root X-1 / root x = B, find the value of (1 + x square) / X
[note] root X-1 / root X -- under the root (x-1) divided by the root x = B; (1 + x square) / X -- (1 + x square) divided by X
Sorry, wrong number. It should be: root X-1 / root X - X-1 under root divided by x = B under root; (1 + x square) / X - (1 + x square) divided by X

√x-1/√x=b
Square on both sides
x-2√x*1/√x+1/x=b^2
x-2+1/x=b^2
x+1/x=b^2+2
(1+x^2)/x
Divide by X
=1/x+x
=b^2+2

When x = 1 + 20022, the value of the algebraic expression (4x3-2005x-2001) 2003 is ()
A. 0B. -1C. 1D. -22003

∵ x = 1 + 20022, ∵ 2x-1 = 2002, both sides are squared to get 4x2-4x + 1 = 2002, namely 4x2-4x = 2001, ∵ 4x3-2005x-2001 = 4x3-2005x - (4x2-4x) = 4x3-4x2-2005x + 4x = x (4x2-4x-2001) = 0, ∵ 4x3-2005x-2001) 2003 = 0